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Author:sikander_nsit
Tester:sikander_nsit
Editorialist:sikander_nsit

DIFFICULTY:

MEDIUM

PROBLEM:

In this problem, we had to find the count of numbers between a range[R,L] such that their GCD with N is X.

EXPLANATION:

It would be the same as finding the count of numbers between L/X and R/X such that their GCD with N/X is 1. This is because if divide two numbers by their GCD then the resulting numbers would be coprime. So the task remains to find the count of numbers between a range which are coprime to N.

Let f(C) be the number of integers from 1 to A which are coprime to N.
So we can calculate our answer as f(R)-f(L-1). f(C) is A minus the number of integers that are not relatively prime to N. Call this number g(C). So f(C)=C−g(C). We attack the problem of finding g(C).

If N is a prime power p^a, it is easy.
The numbers in the interval [1,C] that are not relatively prime to N are the multiples of p.
Thus

g(C)=⌊C/p⌋

where ⌊x⌋ is the usual "floor" function.

If N has prime power factorization p^a*q^b, where p and q are distinct primes, then g(C) is the number of integers in [1,C] that are divisible by p or q or both. By Inclusion/Exclusion, we obtain

g(C)=⌊C/p⌋+⌊C/q⌋−⌊C/pq⌋.

The reason is that when we add the first two terms above, we are counting twice all the multiples of pq.

If N has prime power factorization p^a*q^b*r^c, the same basic idea works. We get

g(C)=⌊C/p⌋+⌊C/q⌋+⌊C/r⌋−⌊C/qr⌋−⌊C/pr⌋−⌊C/pq⌋+⌊C/pqr⌋.

The number of distinct primes for a number less than 10^9 will not be greater than 9. So bitmasking can be used to calculate all product combinations and then calculating f(R) and f(L-1). Each query can be answered in
O(2^number of distinct primes).

AUTHOR'S SOLUTION:

Author's solution can be found here.